Given: In Trapezium ABCD, AB ∥ DC and AD = BC. Construction: Draw AD ∥ BE. Proof: In Quadrilateral ABCD, AB ∥ DE [∵ Given] AD ∥ BE [∵ By construction] Hence, ABED is a Parallelogram. AD = BE [∵ opposite sides of a parallelogram are equal] AD = BC [∵ Given] ⇒ BE = BC In ΔEBC, BE = BC [∵ Proved above]Read more
Given: In Trapezium ABCD, AB ∥ DC and AD = BC.
Construction: Draw AD ∥ BE.
Proof: In Quadrilateral ABCD,
AB ∥ DE [∵ Given]
AD ∥ BE [∵ By construction]
Hence, ABED is a Parallelogram.
AD = BE [∵ opposite sides of a parallelogram are equal]
AD = BC [∵ Given]
⇒ BE = BC
In ΔEBC, BE = BC [∵ Proved above]
Hence, ∠C = ∠2 …(1) [∵ In an isosceles triangle, the angles opposite to equal sides are equal]
∠A = ∠1 …(2) [∵ Opposite angles of a parallelogram are equal]
Here, ∠1 + ∠2 = 180° [∵ Linear pair]
⇒ ∠A + ∠C = 180° [∵ From the equation (1) and (2)]
⇒ ABED is a cyclic quadrilateral.
Construction: Join AD. Proof: AB is diameter of circle and ∠ADB is Formed in semi-circle. Hence, ∠ADB = 90° ...(1) [∵ Angle in a semicircle is a right angle] Similarly, AC is diametar and ∠ADC is formed in semi- circle. Hence, ∠ADC = 90° ...(2) [∵ Angle in a semicircle is a right angle] Here, ∠ADB +Read more
Construction: Join AD.
Proof: AB is diameter of circle and ∠ADB is Formed in semi-circle.
Hence, ∠ADB = 90° …(1) [∵ Angle in a semicircle is a right angle]
Similarly, AC is diametar and ∠ADC is formed in semi- circle.
Hence, ∠ADC = 90° …(2) [∵ Angle in a semicircle is a right angle]
Here, ∠ADB + ∠ADC = 90° + 90° = 180°
∠ADB and ∠ADC are forming linear pair. Therefore BDC is a straight line.
Hence, the point D lies on third side BC.
Given: Triangle ABC and ACD are two right triangle on common base AC. To prove: ∠CAD = ∠CBD. Proof: Triangle ABC and ADC are on common base BC and ∠BAC = ∠BDC. Hence, points A, B, C and D lie on the same circle. [∵ if a line segment joining two points subtends equal angles at two other points lyingRead more
Given: Triangle ABC and ACD are two right triangle on common base AC.
To prove: ∠CAD = ∠CBD.
Proof: Triangle ABC and ADC are on common base BC and ∠BAC = ∠BDC.
Hence, points A, B, C and D lie on the same circle.
[∵ if a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle.]
Therefore,
∠CAD = ∠CBD
[∵ Angles in the some segment are equal]
Given: Quadrilateral ABCD is a cyclic quadrilateral. To Prove: ABCD is a rectangle. Proof: In cyclic quadrilateral ABCD ∠A + ∠C = 180° ...(1) [∵ The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.] But ∠A = ∠C ...(2) [∵ Opposite angles of a parallelogram are equal] From theRead more
Given: Quadrilateral ABCD is a cyclic quadrilateral.
To Prove: ABCD is a rectangle.
Proof: In cyclic quadrilateral ABCD
∠A + ∠C = 180° …(1)
[∵ The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.]
But ∠A = ∠C …(2)
[∵ Opposite angles of a parallelogram are equal]
From the equation (1) and (2).
∠A + ∠A = 180°
⇒ 2∠A = 180°
⇒ ∠A = 180°/2 = 90°
We know that, a parallelogram with one angle right angle, is a rectangle.
Hence, ABCD is a rectangle.
AC is Diameter of circle. Hence, ∠ADC = 90° and ∠ABC = 90° ...(1) [∵ Angle in a Semicircle is a right angle.] Similarly, BD is Diameter of circle. Hence, ∠BAD = 90° and ∠BCD = 90 ...(2) [∵ Angle in a semicircle is a right angle.] From the equation (1) and (2), ∠ADC = ∠ABC = ∠BAD = ∠BCD = 90° Hence,Read more
AC is Diameter of circle.
Hence, ∠ADC = 90° and ∠ABC = 90° …(1) [∵ Angle in a Semicircle is a right angle.]
Similarly, BD is Diameter of circle.
Hence, ∠BAD = 90° and ∠BCD = 90 …(2) [∵ Angle in a semicircle is a right angle.]
From the equation (1) and (2), ∠ADC = ∠ABC = ∠BAD = ∠BCD = 90°
Hence, ABCD is a rectangle.
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Given: In Trapezium ABCD, AB ∥ DC and AD = BC. Construction: Draw AD ∥ BE. Proof: In Quadrilateral ABCD, AB ∥ DE [∵ Given] AD ∥ BE [∵ By construction] Hence, ABED is a Parallelogram. AD = BE [∵ opposite sides of a parallelogram are equal] AD = BC [∵ Given] ⇒ BE = BC In ΔEBC, BE = BC [∵ Proved above]Read more
Given: In Trapezium ABCD, AB ∥ DC and AD = BC.
See lessConstruction: Draw AD ∥ BE.
Proof: In Quadrilateral ABCD,
AB ∥ DE [∵ Given]
AD ∥ BE [∵ By construction]
Hence, ABED is a Parallelogram.
AD = BE [∵ opposite sides of a parallelogram are equal]
AD = BC [∵ Given]
⇒ BE = BC
In ΔEBC, BE = BC [∵ Proved above]
Hence, ∠C = ∠2 …(1) [∵ In an isosceles triangle, the angles opposite to equal sides are equal]
∠A = ∠1 …(2) [∵ Opposite angles of a parallelogram are equal]
Here, ∠1 + ∠2 = 180° [∵ Linear pair]
⇒ ∠A + ∠C = 180° [∵ From the equation (1) and (2)]
⇒ ABED is a cyclic quadrilateral.
. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Construction: Join AD. Proof: AB is diameter of circle and ∠ADB is Formed in semi-circle. Hence, ∠ADB = 90° ...(1) [∵ Angle in a semicircle is a right angle] Similarly, AC is diametar and ∠ADC is formed in semi- circle. Hence, ∠ADC = 90° ...(2) [∵ Angle in a semicircle is a right angle] Here, ∠ADB +Read more
Construction: Join AD.
See lessProof: AB is diameter of circle and ∠ADB is Formed in semi-circle.
Hence, ∠ADB = 90° …(1) [∵ Angle in a semicircle is a right angle]
Similarly, AC is diametar and ∠ADC is formed in semi- circle.
Hence, ∠ADC = 90° …(2) [∵ Angle in a semicircle is a right angle]
Here, ∠ADB + ∠ADC = 90° + 90° = 180°
∠ADB and ∠ADC are forming linear pair. Therefore BDC is a straight line.
Hence, the point D lies on third side BC.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Given: Triangle ABC and ACD are two right triangle on common base AC. To prove: ∠CAD = ∠CBD. Proof: Triangle ABC and ADC are on common base BC and ∠BAC = ∠BDC. Hence, points A, B, C and D lie on the same circle. [∵ if a line segment joining two points subtends equal angles at two other points lyingRead more
Given: Triangle ABC and ACD are two right triangle on common base AC.
See lessTo prove: ∠CAD = ∠CBD.
Proof: Triangle ABC and ADC are on common base BC and ∠BAC = ∠BDC.
Hence, points A, B, C and D lie on the same circle.
[∵ if a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle.]
Therefore,
∠CAD = ∠CBD
[∵ Angles in the some segment are equal]
Prove that a cyclic parallelogram is a rectangle.
Given: Quadrilateral ABCD is a cyclic quadrilateral. To Prove: ABCD is a rectangle. Proof: In cyclic quadrilateral ABCD ∠A + ∠C = 180° ...(1) [∵ The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.] But ∠A = ∠C ...(2) [∵ Opposite angles of a parallelogram are equal] From theRead more
Given: Quadrilateral ABCD is a cyclic quadrilateral.
See lessTo Prove: ABCD is a rectangle.
Proof: In cyclic quadrilateral ABCD
∠A + ∠C = 180° …(1)
[∵ The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.]
But ∠A = ∠C …(2)
[∵ Opposite angles of a parallelogram are equal]
From the equation (1) and (2).
∠A + ∠A = 180°
⇒ 2∠A = 180°
⇒ ∠A = 180°/2 = 90°
We know that, a parallelogram with one angle right angle, is a rectangle.
Hence, ABCD is a rectangle.
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
AC is Diameter of circle. Hence, ∠ADC = 90° and ∠ABC = 90° ...(1) [∵ Angle in a Semicircle is a right angle.] Similarly, BD is Diameter of circle. Hence, ∠BAD = 90° and ∠BCD = 90 ...(2) [∵ Angle in a semicircle is a right angle.] From the equation (1) and (2), ∠ADC = ∠ABC = ∠BAD = ∠BCD = 90° Hence,Read more
AC is Diameter of circle.
See lessHence, ∠ADC = 90° and ∠ABC = 90° …(1) [∵ Angle in a Semicircle is a right angle.]
Similarly, BD is Diameter of circle.
Hence, ∠BAD = 90° and ∠BCD = 90 …(2) [∵ Angle in a semicircle is a right angle.]
From the equation (1) and (2), ∠ADC = ∠ABC = ∠BAD = ∠BCD = 90°
Hence, ABCD is a rectangle.